# Change of variables for multiple integrals (3)

1. Jan 24, 2008

### kingwinner

Q1: Let S be the region in the first quadrant bounded by the curves xy=1, xy=3, x2 - y2 = 1, and x2 - y2 = 4. Compute
∫∫(x2 + y2)dA.
S
(Hint: Let G(x,y)=(xy, x2 - y2). What is |det DG|?)

Solution:

I don't understand the third and fourth equalities in the solutions. Can someone kindly explain?

2. Jan 24, 2008

### HallsofIvy

Staff Emeritus
If you don't understand why
$$\int\int_S(x^2+ y^2)dA= \frac{1}{2}\int\int_S |det DG| dA$$
then you have missed the whole point of this section! |det DG| is the "Jacobian" and |det DG| dA is the differential in terms of the new variables u= xy, v= x2- y2.
[tex]det DG= \left[\begin{array}{cc}\frac{\partial xy}{\partial x} & \frac{\partial x^2- y^2}{\partial x} \\ \frac{\partial xy}{\partial y} & \frac{\partial x^2- y^2}{\partial y}\end{array}\right]= \left[\begin{array}{cc}y & 2x \\ x & -2y\end{array}\right]= -2(x^2+ y^2)[/itex]
as they say so |det dG|= 2(x2+ y2). You already have x2+ y2 in the integral so you only need to multiply inside the integral by 2 and multiply outside the integral by 1/2 to get $1/2 \int\int |det dG|dA$.

Now, for the limits of integration. Your region is bounded by xy= 1, xy= 3, x2- y2= 1 and x2- y2= 4. That, of course, was the reason for the choice of G. With u= xy, u= xy= 1 and u= xy= 3 are the u limits of integration. With v= x2- y2, v= x2- y2= 1 and v= x2- y2= 4 are the v limits of integration.

3. Jan 25, 2008

### kingwinner

Change of variables theorem says
∫∫f(x,y)dA
S
=∫∫f(G(u))|det DG| dudv ?
G-1(S)

So you should add the Jaconbian factor, but why are they moving it away? (it seems that they are doing it the oppposite way than the theorem states)

For the fourth equality, I don't understand why |det DG| disappeared.

4. Jan 25, 2008

### HallsofIvy

Staff Emeritus
You are going in the wrong direction.
Suppose you had the problem $\int x dx$ (yes, I know that's trivial!) and you decide to make the substitution u= x2 (yes, I know that's a peculiar choice!). Then du= 2x dx so (1/2)du= xdx. Since you already have "xdx" in the integral, in terms of u, you have just $(1/2)\int du= (1/2)u+ C= (1/2)x^2+ C$. The "x" inside the integral was "absorbed" into the du.